13. [aqueous Solutions] Practice

1. e.) According to the solubility curve shown here, how many grams of sodium chloride could you dissolve in 100 mL of water at 10 degrees?

35 g

10 g

30 g

80 g

Based on the solubility curve shown, at 10 degrees, the maximum amount of sodium chloride that can be dissolved in 100 mL of water is 35 g. This means that if you were to add more than 35 g of sodium chloride to the 100 mL of water at 10 degrees, it would not dissolve and would form a precipitate.

Explanation

Based on the solubility curve shown, at 10 degrees, the maximum amount of sodium chloride that can be dissolved in 100 mL of water is 35 g. This means that if you were to add more than 35 g of sodium chloride to the 100 mL of water at 10 degrees, it would not dissolve and would form a precipitate.

2. a.) Why doesn't vegetable oil dissolve in water?

Water is a polar molecule while oil is nonpolar.

Water is a nonpolar molecule while oil is polar.

Both molecules are nonpolar.

Both molecules are polar.

The polar nature of water causes water molecules to have strong forces of attraction toward each other. This means that water molecules would rather stick together than mix with vegetable oil. In other words, vegetable oil has no polar sites to which water can attach.

Explanation

The polar nature of water causes water molecules to have strong forces of attraction toward each other. This means that water molecules would rather stick together than mix with vegetable oil. In other words, vegetable oil has no polar sites to which water can attach.

3. d.) Which of the following happens when salt is added to water?

The boiling point increases.

The freezing point decreases.

The vapor pressure decreases.

All of the above.

Recall that each of these observations (along with osmotic pressure) are known as the colligative properties of a solution. The more salt you dissolve, the stronger water's intermolecular forces become (due to attraction to the full ionic charges of salt ions), making the solution prefer the liquid phase.

Explanation

Recall that each of these observations (along with osmotic pressure) are known as the colligative properties of a solution. The more salt you dissolve, the stronger water's intermolecular forces become (due to attraction to the full ionic charges of salt ions), making the solution prefer the liquid phase.

4. b.) To help dissolve a solute, we can add heat. This tells us that the process of dissolving a solute is...

Endothermic.

Exothermic.

Isothermic.

Remember, in an endothermic process, energy flows into the sample; exothermic means energy exits the sample. You can always think about the opposite process -- when the sodium acetate precipitated (un-dissolved), it released heat.

Explanation

Remember, in an endothermic process, energy flows into the sample; exothermic means energy exits the sample. You can always think about the opposite process -- when the sodium acetate precipitated (un-dissolved), it released heat.

5. c.) Which of the following is the BEST electrolyte?

Salt water

Sugar water

Pure water

Carbonated water

Recall that electrolytes are ions that are dissolved in solution. When ionic compounds dissolve, the ions separate from each other. This enables the solution to conduct electricity much better than fresh water.

Explanation

Recall that electrolytes are ions that are dissolved in solution. When ionic compounds dissolve, the ions separate from each other. This enables the solution to conduct electricity much better than fresh water.

6. f.) Tom dissolves 20 grams of potassium chloride in 100 mL of water at 30 degrees. According to the solubility curve, this solution can best be described as...

Unsaturated.

Saturated.

Supersaturated.

Remember that the line in a solubility curve represents the saturation point where no more solute will dissolve under those conditions. The area under the curve represents an unsaturated solution, while a supersaturated solution would be above the curve. Don't forget -- in order to create a supersaturated solution, a heated, saturated solution must be cooled!

Explanation

Remember that the line in a solubility curve represents the saturation point where no more solute will dissolve under those conditions. The area under the curve represents an unsaturated solution, while a supersaturated solution would be above the curve. Don't forget -- in order to create a supersaturated solution, a heated, saturated solution must be cooled!

7. h.) If Harriet wants to create 250 mL of a 0.5 M CO2 solution, how many grams of CO2 must she dissolve?

125 g

2.8 g

0.55 g

0.125 g

This is just like standard (i), only backwards. We know that molarity = moles / liters. To find the number of moles of solute we need to use, we plug in the values we know.
0.5 M = moles CO2 / 0.25 L
=0.125 moles CO2

We must then convert moles into grams by multiplying by the molar mass (44 g/mol), giving us 0.55 g.

Explanation

This is just like standard (i), only backwards. We know that molarity = moles / liters. To find the number of moles of solute we need to use, we plug in the values we know.
0.5 M = moles CO2 / 0.25 L
=0.125 moles CO2

We must then convert moles into grams by multiplying by the molar mass (44 g/mol), giving us 0.55 g.

8. i.) 200 mL of water were added to 150 mL of a 2 M Kool Aid solution. What is the new concentration?

0.86 M

1.5 M

0.01 M

2.7 M

This is a dilution problem, so we must use the dilution equation:
M1 x V1 = M2 x V2
(2M) x (150 mL) = (M2) x (350 mL)
M2 = 0.86 M
The part that is most often missed is the fact that V2 is the TOTAL final volume. Since we added 200 mL to the 150 mL we already had, we have a total of 350 mL. Also note that it doesn't matter which units we use for volume as long as it's the same unit on both sides of the equation.

Explanation

This is a dilution problem, so we must use the dilution equation:
M1 x V1 = M2 x V2
(2M) x (150 mL) = (M2) x (350 mL)
M2 = 0.86 M
The part that is most often missed is the fact that V2 is the TOTAL final volume. Since we added 200 mL to the 150 mL we already had, we have a total of 350 mL. Also note that it doesn't matter which units we use for volume as long as it's the same unit on both sides of the equation.

9. g.) If I dissolve 20 grams of sodium chloride in 100 mL of water, what is the concentration of my solution?

5 M

0.2 M

0.34 M

3.4 M

We calculate concentration in terms of molarity, or moles/Liter. To find the concentration, we need to know the number of moles of solute and the number of liters of solvent.
Starting with the solute, we know we have 20 grams of NaCl. To convert into moles, we divide by the molar mass (58.5 g/mol) to give us 0.34 moles.
We then divide by the number of liters of solvent. Since we are given mL as our value, we must slide the decimal three places to the left, giving us 0.1 L.
0.34 moles / 0.1 L = 3.4 M

Explanation

We calculate concentration in terms of molarity, or moles/Liter. To find the concentration, we need to know the number of moles of solute and the number of liters of solvent.
Starting with the solute, we know we have 20 grams of NaCl. To convert into moles, we divide by the molar mass (58.5 g/mol) to give us 0.34 moles.
We then divide by the number of liters of solvent. Since we are given mL as our value, we must slide the decimal three places to the left, giving us 0.1 L.
0.34 moles / 0.1 L = 3.4 M